Prove the following identities without the use of a calculator: $$ \begin{array}{ll} ext { a) } \cos \left( heta+60^{\circ} ight)-\sqrt{3} \cos \left( heta+30^{\circ} ight)=-\cos heta & ext { d) } \cos 2 heta=1-2 \sin ^{2} heta \ ext { b) } \cos (\alpha+\beta)+\cos (\alpha-\beta)=2 \cos \alpha \cos \beta & ext { e) } \sin \left(x+30^{\circ} ight)-\sin \left(x-30^{\circ} ight)=\cos x \ ext { c) } \sin 2 heta=2 \sin heta \cos heta & ext { f) } \cos 7 \mathrm{~A}+\cos \mathrm{A}=2 \cos 4 \mathrm{~A} \cos 3 \mathrm{~A}\end{array} $$

Question

Prove the following identities without the use of a calculator: $$ \begin{array}{ll}	ext { a) } \cos \left(	heta+60^{\circ}ight)-\sqrt{3} \cos \left(	heta+30^{\circ}ight)=-\cos 	heta & 	ext { d) } \cos 2 	heta=1-2 \sin ^{2} 	heta \ 	ext { b) } \cos (\alpha+\beta)+\cos (\alpha-\beta)=2 \cos \alpha \cos \beta & 	ext { e) } \sin \left(x+30^{\circ}ight)-\sin \left(x-30^{\circ}ight)=\cos x \ 	ext { c) } \sin 2 	heta=2 \sin 	heta \cos 	heta & 	ext { f) } \cos 7 \mathrm{~A}+\cos \mathrm{A}=2 \cos 4 \mathrm{~A} \cos 3 \mathrm{~A}\end{array} $$

UpStudy AI · Accepted Answer

**a) Prove that**  
$$
\cos\left(\theta+60^\circ\right)-\sqrt{3}\cos\left(\theta+30^\circ\right)=-\cos\theta.
$$

1. Use the cosine addition formulas:
   - $$\cos\left(\theta+60^\circ\right)=\cos\theta\cos60^\circ-\sin\theta\sin60^\circ=\cos\theta\left(\frac{1}{2}\right)-\sin\theta\left(\frac{\sqrt{3}}{2}\right)$$.
   - $$\cos\left(\theta+30^\circ\right)=\cos\theta\cos30^\circ-\sin\theta\sin30^\circ=\cos\theta\left(\frac{\sqrt{3}}{2}\right)-\sin\theta\left(\frac{1}{2}\right)$$.

2. Substitute into the given expression:
   $$
   \begin{aligned}
   &\cos\left(\theta+60^\circ\right)-\sqrt{3}\cos\left(\theta+30^\circ\right)\[1mm]
   &= \left[\frac{1}{2}\cos\theta-\frac{\sqrt{3}}{2}\sin\theta\right]-\sqrt{3}\left[\frac{\sqrt{3}}{2}\cos\theta-\frac{1}{2}\sin\theta\right].
   \end{aligned}
   $$

3. Distribute and simplify:
   $$
   \begin{aligned}
   &= \frac{1}{2}\cos\theta-\frac{\sqrt{3}}{2}\sin\theta-\frac{3}{2}\cos\theta+\frac{\sqrt{3}}{2}\sin\theta\[1mm]
   &= \left(\frac{1}{2}-\frac{3}{2}\right)\cos\theta + \left(-\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}\right)\sin\theta\[1mm]
   &= -\cos\theta.
   \end{aligned}
   $$

---

**b) Prove that**  
$$
\cos(\alpha+\beta)+\cos(\alpha-\beta)=2\cos\alpha\cos\beta.
$$

1. Apply the cosine addition formulas:
   $$
   \cos(\alpha+\beta)=\cos\alpha\cos\beta - \sin\alpha\sin\beta,
   $$
   $$
   \cos(\alpha-\beta)=\cos\alpha\cos\beta + \sin\alpha\sin\beta.
   $$

2. Sum the two expressions:
   $$
   \cos(\alpha+\beta)+\cos(\alpha-\beta) = \left(\cos\alpha\cos\beta - \sin\alpha\sin\beta\right)+\left(\cos\alpha\cos\beta+\sin\alpha\sin\beta\right).
   $$

3. Combine like terms:
   $$
   = 2\cos\alpha\cos\beta.
   $$

---

**c) Prove that**  
$$
\sin2\theta=2\sin\theta\cos\theta.
$$

1. Recall the standard double-angle formula for sine:
   $$
   \sin2\theta = 2\sin\theta\cos\theta.
   $$

This is a well-known identity by definition.

---

**d) Prove that**  
$$
\cos2\theta=1-2\sin^2\theta.
$$

1. Use one of the double-angle formulas for cosine:
   $$
   \cos2\theta=\cos^2\theta-\sin^2\theta.
   $$

2. Express $$\cos^2\theta$$ in terms of $$\sin^2\theta$$ using the Pythagorean identity:
   $$
   \cos^2\theta=1-\sin^2\theta.
   $$

3. Substitute into the double-angle formula:
   $$
   \cos2\theta=(1-\sin^2\theta)-\sin^2\theta=1-2\sin^2\theta.
   $$

---

**e) Prove that**  
$$
\sin\left(x+30^\circ\right)-\sin\left(x-30^\circ\right)=\cos x.
$$

1. Use the sine subtraction formula:
   $$
   \sin u-\sin v=2\cos\left(\frac{u+v}{2}\right)\sin\left(\frac{u-v}{2}\right).
   $$
   Here, let $$u=x+30^\circ$$ and $$v=x-30^\circ$$.

2. Compute the averages:
   $$
   \frac{u+v}{2}=\frac{(x+30^\circ)+(x-30^\circ)}{2}= \frac{2x}{2}=x,
   $$
   $$
   \frac{u-v}{2}=\frac{(x+30^\circ)-(x-30^\circ)}{2}= \frac{60^\circ}{2}=30^\circ.
   $$

3. Substitute back:
   $$
   \sin\left(x+30^\circ\right)-\sin\left(x-30^\circ\right)=2\cos x\sin30^\circ.
   $$

4. Since $$\sin30^\circ=\frac{1}{2}$$, then:
   $$
   2\cos x\left(\frac{1}{2}\right)=\cos x.
   $$

---

**f) Prove that**  
$$
\cos7A+\cos A=2\cos4A\cos3A.
$$

1. Use the sum-to-product formula for cosine:
   $$
   \cos u+\cos v=2\cos\left(\frac{u+v}{2}\right)\cos\left(\frac{u-v}{2}\right).
   $$
   Here, set $$u=7A$$ and $$v=A$$.

2. Compute the averages:
   $$
   \frac{u+v}{2}=\frac{7A+A}{2}=4A,
   $$
   $$
   \frac{u-v}{2}=\frac{7A-A}{2}=3A.
   $$

3. Substitute into the formula:
   $$
   \cos7A+\cos A=2\cos4A\cos3A.
   $$

Each identity has been proven step by step using standard trigonometric formulas.
**a)** $$ \cos\left(\theta+60^\circ\right)-\sqrt{3}\cos\left(\theta+30^\circ\right) = -\cos\theta $$ **b)** $$ \cos(\alpha+\beta) + \cos(\alpha-\beta) = 2\cos\alpha\cos\beta $$ **c)** $$ \sin2\theta = 2\sin\theta\cos\theta $$ **d)** $$ \cos2\theta = 1 - 2\sin^2\theta $$ **e)** $$ \sin\left(x+30^\circ\right) - \sin\left(x-30^\circ\right) = \cos x $$ **f)** $$ \cos7A + \cos A = 2\cos4A\cos3A $$ All identities have been proven using trigonometric formulas.

Prove the following identities without the use of a calculator:

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